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chapter-1
1. Chapter 1
Motion in a straight line
1-2
Displacement vs Distance Average Velocity
1-3 INSTANTANEOUS VELOCITY
1 - 4 Acceleration
1.6 the acceleration of gravity and falling objects
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2. 1-2 Displacement vs Distance
Average Velocity
• Displacement is a vector that points from an object’s
initial position to its final position
• and has a magnitude that equals the shortest distance
between the two positions.
_Only depends on the initial and final positions
– Independent of actual paths between the initial and
final positions
• Distance is a scalar
– Depends on the initial and final positions as well as
the actual path between them
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3. Displacement
This type of x(t) plot shows the position
of an object at any time, e.g.,
Position at t=3 s, x(3) = 1 m
x (m)
3
4 t (s)
Displacement between t=1 s and t=5 s
∆x = 1.0 m - 2.0 m = -1.0 m
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4. Given the train’s initial position and its final position what is the
displacement of the train?
What is the distance traveled by the train ?
Displacement = ∆x = x − x
f
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5. Example:
A boy travels from D to A,A to B .B to C.C to D
Displacement from D to D ( which are initial and final points ) = 0
Distance traveled = 8 +4+8+4 = 24 m
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6. Example :
Distance = 4 m + 3 m =7 m
Displacement = 5 m
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7. Speed and Velocity
The average speed being the distance traveled
divided by the time required to cover the distance:
How far does a jogger run in 1.5 hours (5400 s) if his average
speed is 2.22 m /s?
Distance = 5400 s x 2.22 m / s = 11988
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8. Speed
Speed can be defined in a couple of ways:
How fast something is moving
The distance covered in a certain amount of time
The rate of change of the position of an object
Units for speed are:
This is the standard unit
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miles / hour (mi/hr)
kilometers / hour (km/hr)
feet / second (ft/s)
meters / second (m/s)
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11. example
A particle moves along a straight line such that its
position is defined by s = (t3 – 3 t2 + 2 ) m.
Determine the velocity of the particle when t = 4 s.
dx
v=
= 3t 2 − 6t
dt
At
t = 4 s,
the velocity = 3 (4)2 – 6(4) = 24 m/s
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17. 1-3 INSTANTANEOUS
VELOCITY
Instantaneous velocity – is how fast an object is moving at a particular
instant.
∆
x
dx
example
v = lim
∆→
t
0
∆
t
=
dt
The position of a particle moving on the x axis is given by
x = 7.8 + 9.2 t – 2.1t2. What is its instantaneous velocity at t = 3.5 seconds
v = 0 + 9.2 – (2)(2.1)t
v = 0 + 9.2 – (2)(2.1)(3.5) = -5.5 m/s
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18. 1 - 4 Acceleration
Acceleration: is a rate at which a velocity is changing.
Instantaneous acceleration
= dv / dt = d2 x / d t2
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19. Example
A car’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the
bottom of the hill with a velocity of 26 m/s. What is the acceleration of the
car?
The car is increasing its velocity by 8 m/s for every second
it is moving.
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21. • Instantaneous Acceleration
• Instantaneous Acceleration
∆v dv
a = lim
=
∆t →0 ∆t
dt
• Suppose a particle is moving in a straight line so that its position is
given by the relationship x = (2.10 m/s2)t3 + 2.8 m. Find its
instantaneous acceleration at 5 seconds.
v = dx / dt = (3)(2.1)t2
a = dv / dt = (2)(3)(2.1)t
at
t= 5s
a = (2)(3)(2.1)(5) = 63 m/s2
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23. example
A bullet train starts from rest from a station and travels along a straight
horizontal track towards another station. The graph in fig. shows
how the speed of the train varies withtime over the whole journey.
Determine: (a) the total distance covered by the train,
(b) the average speed of the train.
A)
Total distance travelled
40 + 0
40 + 0
x=
× 2 + 40 × 10 +
×4
2
2
x = 40 + 400 + 80 = 520m
OR
Total distance travelled
= ‘area under the graph’
= (1/2)(10 + 16)(40) = 520 m
Speed / ms-1
40
0
Time
2
12 16
Average speed = (total distance) / (total time ) = 520 / 16 = 32.5 ms-1
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24. example
: a car is traveling 30 m/s and approaches 10 m from an
intersection when the driver sees a pedestrian and slams
on his brakes and decelerates at a rate of 50 m/s2.
(a) How long does it take the car to come to a stop?
(b) how far does the car travel before coming to a stop?
vf -vi= a t, where vo= 30 m/s, v = 0 m/s, and a = -50 m/s 2
t = (0 -30)/(-50) = 0.6 s
Δx= vit + ½ a t2= (30)(0.6) + ½(-50)(0.6)2= 18 -9 = 9 m
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25. 1.5 finding the motion of an object
Equations of Kinematics for Constant Acceleration
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27. example
- A car starting from rest attains a speed of 28 m/sec in 20 sec. Find the
acceleration of the car and the distance it travels in this time.
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29. 1. Velocity & acceleration are both vectors.
Are the velocity and the acceleration always in the same
direction?
NO!!
If the object is slowing down, the acceleration
vector is in the opposite direction of the velocity
vector!
2. Velocity & acceleration are vectors.
Is it possible for an object to have a zero acceleration and a
non-zero velocity?
YES!!
If the object is moving at a constant velocity, the
acceleration vector is zero!
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31. Examples :
1 ) What is the acceleration of a car that increased its speed from 10
m/s to 30 m/s in 4 seconds?
a = (30 m/s – 10 m/s) ÷ 4s
=
20 m/s ÷ 4s
=
5 m/s2
2)the same car now slows down back to 10 m/s in 5 seconds. What is
his acceleration?
a = (10 m/s – 30 m/s) ÷ 5s
=
(- 20 m/s) ÷ 5s
=
- 4 m/s2
Means slowing down
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32. Graphical Analysis
*deduce from the shape of a speed-time graph when a body is:
(i) at rest
(ii) moving with uniform speed
(iii) moving with uniform acceleration
(iv) moving with non-uniform acceleration
Velocity
(ii)
(iii)
(i) at rest
(ii) moving with uniform speed
(iii) moving with uniform
acceleration
(iv) moving with non-uniform
acceleration
(iv)
(i)
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Time
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34. A bus stopped at a bus-stop for 10 seconds before accelerating to a velocity
of 15 m/s in 4 seconds and then at a constant speed for the next 9 seconds.
How does the graph look like? How far did the bus go in this 23 seconds?
• Distance travelled in first 10 seconds is zero
• Distance travelled in the next 4 seconds is
= ½ x 4 x 15 = 30 m
Velocity /m/s
• Distance travelled in the final 9 seconds is
(ii) = 9 x 15 = 135 m
15
• Total distance travelled = 165 m
(iii)
(i)
0
10
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Time/s
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36. 1.6 the acceleration of gravity and falling objects
Objects thrown straight up
The acceleration of a falling object is due to the force of
gravity between the object and the earth.
Galileo showed that falling objects accelerate equally,
neglecting air resistance.
Galileo found that all things fall at the same rate.
On the surface of the earth, in a vacuum, all objects
accelerate towards the surface of the earth at 9.8 m/s2.
The acceleration of gravity (g) for objects in free fall at
the earth's surface is 9.8 m/s2. ( down ward )
g actually changes as we move to higher altitudes
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37. Equations of Kinematics for Constant Acceleration
For free fall
v = v0 + at
v = v0 − gt
1 2
∆x = v0t + at
2
2
2
v = v0 + 2a∆x
1 2
∆y = v0t − gt
2
2
v 2 = v0 − 2g∆y
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39. A ball is dropped from a tall building and strikes the ground 4
seconds later.
A ) what velocity does it strike the ground
B ) what distance does it fall?
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45. Word clues to numbers for problem
solving
• “free-fall” acceleration due to gravity
a=9.8m/s2, down
• “at rest” not moving
v=0
• “dropped” starts at rest and free-fall
vi=0 and a=9.81m/s2, down
• “constant velocity” no acceleration
a=0
• “stops” final velocity is zero
vf=0
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46. Summary
∆x = x f − xi
1.Displacement:
∆x
2. Average velocity: v =
∆t
v f − vi
4. Average
a=
acceleration
∆t
:
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2. Time interval:
3.Instantaneou
s velocity:
v = Lim∆t →0
5.nstantaneous
acceleration:
Norah Ali Almoneef
∆x dx
=
∆t dt
a = Lim∆t →0
∆v dv
=
∆t dt
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